5.1一维波动方程习题参考解答

zhangzujin

# 一维波动方程习题参考解答 --- 1、 求方程 \begin\{aligned\} x^2u\_\{xx\}-y^2u\_\{yy\}-2yu\_y=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的通解. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 特征方程为 \begin\{aligned\} x^2\mathrm\{ d\} y^2-y^2\mathrm\{ d\} x^2=0 \Rightarrow \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}=\pm \frac\{y\}\{x\} \Rightarrow xy=C\_1\mbox\{ 或 \}\frac\{y\}\{x\}=C\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 作自变量变换 \begin\{aligned\} \xi=xy,\quad \eta=\frac\{y\}\{x\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则原方程化为 \begin\{aligned\} 2\eta \frac\{\partial^2u\}\{\partial \xi\partial \eta\} +\frac\{\partial u\}\{\partial \xi\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 令 $\displaystyle \frac\{\partial u\}\{\partial \xi\}=v$, 则 \begin\{aligned\} 2\eta \frac\{\partial v\}\{\partial \eta\}+v=0 \Rightarrow v=\frac\{\phi(\xi)\}\{\sqrt\{\eta\}\} \Rightarrow u=\frac\{\varPhi(\xi)\}\{\sqrt\{\eta\}\}+\varPsi(\eta) =\sqrt\{\frac\{x\}\{y\}\}\varPhi(xy) +\varPsi\left(\frac\{y\}\{x\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 2、 求解下列定解问题: (1)、 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{xx\}+2u\_\{xy\}-3u\_\{yy\}=0,&x\in\mathbb\{R\},\ y > y\_0\\\\ u|\_\{y=y\_0\}=3x^2,\quad u\_y|\_\{y=y\_0\}=0,&x\in\mathbb\{R\}. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 特征方程为 \begin\{aligned\} &0=\mathrm\{ d\} y^2-2\mathrm\{ d\} x\mathrm\{ d\} y-3\mathrm\{ d\} x^2 =(\mathrm\{ d\} y-3\mathrm\{ d\} x)(\mathrm\{ d\} y+\mathrm\{ d\} x)\\\\ \Rightarrow& y-3x=C\_1\mbox\{ 或 \}y+x=C\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 作自变量变换 \begin\{aligned\} \xi=y-3x,\quad \eta =y+x, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则原方程化为 \begin\{aligned\} \frac\{\partial^2u\}\{\partial \xi\partial \eta\}=0 \Rightarrow u=f(\xi)+g(\eta) =f(y-3x)+g(y+x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又由初始条件, \begin\{aligned\} u|\_\{y=y\_0\}=f(y\_0-3x)+g(y\_0+x)&=3x^2,\\\\ u\_y|\_\{y=y\_0\}=f'(y\_0-3x)+g'(y\_0+x)&=0,\\\\ -\frac\{1\}\{3\} f(y\_0-3x)+g(y\_0+x)&=C,\\\\ f(y\_0-3x)&=\frac\{3\}\{4\}x^2-\frac\{3\}\{4\}C,\\\\ g(y\_0+x)&=\frac\{3\}\{4\}x^2+\frac\{3\}\{4\}C,\\\\ u=f(y-3x)+g(y+x)&=3x^2+(y-y\_0)^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{xx\}+2\cos xu\_\{xy\}-\sin^2x u\_\{yy\}-\sin xu\_y=0,&x\in\mathbb\{R\},\ y > \sin x,\\\\ u|\_\{y=\sin x\}=\varphi(x),\ u\_y|\_\{y=\sin x\}=\psi(x),&x\in\mathbb\{R\}. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 特征方程为 \begin\{aligned\} 0=&\mathrm\{ d\} y^2-2\cos x\mathrm\{ d\} x\mathrm\{ d\} y -\sin^2x\mathrm\{ d\} x^2\\\\ =&[\mathrm\{ d\} y-(\cos x-1)\mathrm\{ d\} x][\mathrm\{ d\} y-(\cos x+1)\mathrm\{ d\} x], \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而 \begin\{aligned\} y-\sin x+x=C\_1\mbox\{ 或 \}y-\sin x-x=C\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 作自变量变换 \begin\{aligned\} \xi=y-\sin x+x,\quad \eta=y-\sin x-x, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则原方程化为 \begin\{aligned\} u\_\{\xi\eta\}=0\Rightarrow u=f(\xi)+g(\eta) =f(y-\sin x+x)+g(y-\sin x-x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又由初始条件, \begin\{aligned\} f(x)+g(-x)&=\varphi(x),\\\\ f'(x)+g'(-x)&=\psi(x),\\\\ f(x)-g(x)&=\int\_0^x \psi(t)\mathrm\{ d\} t+C,\\\\ f(x)&=\frac\{1\}\{2\}\left\[\varphi(x)+\int\_0^x \psi(t)\mathrm\{ d\} t+C\right\],\\\\ g(x)&=\frac\{1\}\{2\}\left\[\varphi(-x)+\int\_0^\{-x\}\psi(t)\mathrm\{ d\} t-C\right\],\\\\ u(x,y)&=\frac\{1\}\{2\}[\varphi(y-\sin x+x)+\varphi(-y+\sin x+x)]\\\\ &\quad +\frac\{1\}\{2\}\int\_\{-y+\sin x+x\}^\{y-\sin x+x\} \psi(t)\mathrm\{ d\} t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}-a^2u\_\{xx\} +2u\_t+u=0,&x\in\mathbb\{R\},\ t > 0,\\\\ u|\_\{t=0\}=0,\ u\_t|\_\{t=0\}=x,&x\in\mathbb\{R\}. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle v(x,t)=\mathrm\{e\}^t u(x,t)$, 则 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} v\_\{tt\}-a^2v\_\{xx\}=0, x\in \mathbb\{R\}, t > 0,\\\\ v|\_\{t=0\}=0, v\_t|\_\{t=0\}=x.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} v(x,t)=&\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\xi \mathrm\{ d\} \xi=tx. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 $\displaystyle u(x,t)=tx\mathrm\{e\}^\{-t\}$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (4)、 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} y^2u\_\{xy\}+u\_\{yy\} -\frac\{2\}\{y\}u\_y=0,&x\in\mathbb\{R\},\ y > 1,\\\\ u|\_\{y=1\}=1-x,\ u\_y|\_\{y=1\}=3,&x\in\mathbb\{R\}. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 特征方程为 \begin\{aligned\} 0=-y^2\mathrm\{ d\} x\mathrm\{ d\} y+\mathrm\{ d\} x^2=-\mathrm\{ d\} x(y^2\mathrm\{ d\} y-\mathrm\{ d\} x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 解得 $\displaystyle x=C, x=\frac\{y^3\}\{3\}+C$. 故设 $\displaystyle \xi=x, \eta=x-\frac\{y^3\}\{3\}$ 后, \begin\{aligned\} 0=y^2u\_\{xy\}+u\_\{yy\}-\frac\{2\}\{y\}u\_y=-y^4u\_\{\xi\eta\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 \begin\{aligned\} u\_\{\xi\eta\}=0\Rightarrow u\_\xi=\phi(\xi)\Rightarrow u=f(\xi)+g(\eta) =f(x)+g\left(x-\frac\{y^3\}\{3\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又由 \begin\{aligned\} 1-x=u(x,1)=f(x)+g\left(x-\frac\{1\}\{3\}\right), 3=u\_y(x,1)=g'\left(x-\frac\{1\}\{3\}\right)\cdot(-1) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 \begin\{aligned\} &g'\left(x-\frac\{1\}\{3\}\right)=-3\Rightarrow g\left(x-\frac\{1\}\{3\}\right)=-3x+C\\\\ \Rightarrow& f(x)=1-x+3x-C=1+2x-C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 \begin\{aligned\} u(x,y)=&f(x)+g\left(x-\frac\{y^3\}\{3\}\right)\\\\ =&1+2x-C-3\left(x-\frac\{y^3\}\{3\}+\frac\{1\}\{3\}\right)+C=y^3-x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 3、 设 $\displaystyle A,B,C,D$ 为平面上由特征线所围成的平行四边形的四个顶点, $\displaystyle u$ 为齐次弦振动方程的解, 证明: \begin\{aligned\} u(A)+u(C)=u(B)+u(D). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设弦振动方程为 $\displaystyle u\_\{tt\}-a^2u\_\{xx\}=0$, 而 \begin\{aligned\} A (x,t), B (x-a\eta, t-\eta), C (x+a\xi-a\eta, t-\xi-\eta), D (x+a\xi,t-\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则由 \begin\{aligned\} u(x,t)=\frac\{1\}\{2\}[\varphi(x-at)+\varphi(x+at)] +\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 \begin\{aligned\} u(A)+u(C)=&\frac\{1\}\{2\}\left\[\varphi(x-at)+\varphi(x+at)\right\]+\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\psi(\tau)\mathrm\{ d\}\tau\\\\ &+\frac\{1\}\{2\}[\varphi(x-at+2a\xi)+\varphi(x+at-2a\eta)]\\\\ &+\frac\{1\}\{2a\}\int\_\{x-at+2a\xi\}^\{x+at-2a\eta\} \psi(\tau)\mathrm\{ d\} \tau, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} u(B)+u(D)=&\frac\{1\}\{2\}[\varphi(x-at)+\varphi(x+at-2a\eta)] +\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at-2a\eta\}\psi(\tau)\mathrm\{ d\} \tau\\\\ &+\frac\{1\}\{2\}[\varphi(x-at+2a\xi)+\varphi(x+at)] +\frac\{1\}\{2a\}\int\_\{x-at+2a\xi\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 注意到 \begin\{aligned\} \int\_\{x-at\}^\{x+at\}+\int\_\{x-at+2a\xi\}^\{x+at-2a\eta\} =&\left\[\int\_\{x-at\}^\{x+at\}+\int\_\{x+at\}^\{x-at+2a\xi\}+\int\_\{x-at+2a\xi\}^\{x+at-2a\eta\}\right\] +\int\_\{x-at+2a\xi\}^\{x+at\}\\\\ =&\int\_\{x-at\}^\{x+at-2a\eta\}+\int\_\{x-at+2a\xi\}^\{x+at\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 我们即知结论成立. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 4、 写出 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-4u\_\{xx\}=0, x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=\varphi(x), u\_t(x,0)=\psi(x), x\in\mathbb\{R\}\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 关于点 $\displaystyle (x,t)=(3,5)$ 的依赖区间, 区间 $\displaystyle [8,10]$ 的决定区域, 以及区间 $\displaystyle [1,6]$ 的影响区域. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 点 $\displaystyle (x,t)=(3,5)$ 的依赖区间为 \begin\{aligned\} \left\[x-at,x+at\right\]=[-7,13]; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 区间 $\displaystyle [x\_1,x\_2]=[8,10]$ 的决定区域为 \begin\{aligned\} &\left\\{(x,t); x\_1+at\leq x\leq x\_2-at, t\geq 0\right\\}\\\\ =&\left\\{(x,t); 8+2t\leq x\leq 10-2t, 0\leq t\leq \frac\{1\}\{2\}\right\\}; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 区间 $\displaystyle [x\_1,x\_2]=[1,6]$ 的影响区域为 \begin\{aligned\} \left\\{(x,t); x\_1-at\leq x\leq x\_2+at, t\geq 0\right\\} =\left\\{(x,t); 1-2t\leq x\leq 6+2t, t\geq 0\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 5、 在上半平面 $\displaystyle \left\\{(x,t); x\in\mathbb\{R\}, t > 0\right\\}$ 上给定一点 $\displaystyle M (2,3)$, 对于弦振动方程 $\displaystyle u\_\{tt\}=u\_\{xx\}$ 来说, 点 $\displaystyle M$ 的依赖区间是什么? 点 $\displaystyle M$ 是否落在点 $\displaystyle (1,0)$ 的影响区域内? [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 点 $\displaystyle M$ 的依赖区间是 $\displaystyle [2-3, 2+3]=[-1,5]$. 由 $\displaystyle (1,0)$ 的影响区域为 \begin\{aligned\} 1-t\leq x\leq 1+t, t > 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle t=3\Rightarrow -2\leq x\leq 4$, 而点 $\displaystyle M$ 落在点 $\displaystyle (1,0)$ 的影响区域内. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 6、 试证明 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}-u\_\{xx\}=6(x+t),&x\in\mathbb\{R\},\ t > x,\\\\ u|\_\{t=x\}=0,\ u\_t|\_\{t=x\}=u\_1(x),&x\in\mathbb\{R\}, \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 有解的充分必要条件上 $\displaystyle u\_1(x)-3x^2=$ 常数, 如有解, 解不唯一. 试问: 若把初值给定在直线 $\displaystyle t=ax$ 上, 为什么在 $\displaystyle a=\pm 1$ 与 $\displaystyle a\neq \pm 1$ 的情况, 关于存在唯一性的结论不一样? [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 令 \begin\{aligned\} \xi=x+t,\quad \eta=x-t, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 \begin\{aligned\} -4u\_\{\xi\eta\}=6\xi,\quad u|\_\{\eta=0\}=0,\quad (u\_\xi-u\_\{\eta\})|\_\{\eta=0\}=u\_1\left(\frac\{\xi\}\{2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 如此, \begin\{aligned\} u\_\{\xi\eta\}=-\frac\{3\}\{2\}\xi,\quad u|\_\{\eta=0\} =0,\quad u\_\eta|\_\{\eta=0\}=-u\_1\left(\frac\{\xi\}\{2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 积分有 \begin\{aligned\} u\_\eta=-\frac\{3\}\{4\}\xi^2+\psi(\eta),\quad u=-\frac\{3\}\{4\} \xi^2\eta +\int\_0^\eta \psi(\tau)\mathrm\{ d\} \tau+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由初值条件, \begin\{aligned\} 0=u|\_\{\eta=0\}=C,\quad -u\_1\left(\frac\{\xi\}\{2\}\right) =u\_\{\eta\}|\_\{\eta=0\}=-\frac\{3\}\{4\}\xi^2+C\Rightarrow u\_1(\xi)-3\xi^2=-C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是, 当且仅当 $\displaystyle u\_1(x)-3x^2\equiv $ 常数时, 方程有解, 且为 \begin\{aligned\} &u(\xi,\eta)=-\frac\{3\}\{4\}\xi^2\eta +\int\_0^\eta \psi(\tau)\mathrm\{ d\} \tau\\\\ \Rightarrow&u(x,t)=-\frac\{3\}\{4\}(x-t)(x+t)^2+\int\_0^\{x-t\}\psi(\tau)\mathrm\{ d\} \tau. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} $\displaystyle a=\pm 1$ 与 $\displaystyle a\neq \pm 1$ 时解的存在唯一性结论不一样, 主要在于 $\displaystyle ax-t=0$ 是否为原方程的特征线, 给定不是特征线的曲线的初值, 才能完全确定出二阶导数的初值, 才有存在唯一性. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 7、 求解下列半直线上的混合问题: (1)、 $\displaystyle \left\\{\begin\{array\}\{ll\} u\_\{tt\}-a^2u\_\{xx\}=0, x > 0, t > 0,\\\\ u|\_\{t=0\}=2x,\ u\_t|\_\{t=0\}=x^2, x\geq 0\\\\ u|\_\{x=0\}=0, t\geq 0. \end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \phi(x)=2x, \psi(x)=x^2$, 则对初值作奇延拓知当 $\displaystyle x\geq at$ 时, \begin\{aligned\} u(x,t)=&\frac\{1\}\{2\}[\phi(x-at)+\phi(x+at)] +\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau\\\\ =&\frac\{a^2t^3\}\{3\}+x(2+tx), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 当 $\displaystyle 0\leq x < at$ 时, \begin\{aligned\} u(x,t)=&\frac\{1\}\{2\}[\phi(x+at)-\phi(at-x)]+\frac\{1\}\{2a\}\int\_\{at-x\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau\\\\ =&\frac\{x^3\}\{3a\}+(2+at^2)x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle \left\\{\begin\{array\}\{ll\} u\_\{tt\}-u\_\{xx\}=6t, x > 0, t > 0,\\\\ u|\_\{t=0\}=x^2,\ u\_t|\_\{t=0\}=0, x\geq 0\\\\ u|\_\{x=0\}=t^3, t\geq 0. \end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle v=u-t^3$, 则 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}v\_\{tt\}-v\_\{xx\}=0,&x > 0, t > 0,\\\\ v|\_\{t=0\}=x^2, v\_t|\_\{t=0\}=0, &x\geq 0,\\\\ v|\_\{x=0\}=0,&t\geq 0.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle \phi(x)=x^2, \psi(x)=0$, 则当 $\displaystyle x\geq t$ 时, \begin\{aligned\} u(x,t)=&t^3+\frac\{1\}\{2\}[\phi(x-at)+\phi(x+at)] +\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau\\\\ =&t^2+t^3+x^2; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 当 $\displaystyle 0\leq x < t$ 时, \begin\{aligned\} u(x,t)=&\frac\{1\}\{2\}[\phi(x+at)-\phi(at-x)]+\frac\{1\}\{2a\}\int\_\{at-x\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau =t^3+2tx. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 8、 设 $\displaystyle u(x,t)$ 是下述半无界问题 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}-a^2u\_\{xx\}=0,&0 < x < \infty,\ t > 0,\\\\ u|\_\{t=0\}=\varphi(x),\ u\_t|\_\{t=0\}=\psi(x),&0\leq x < \infty,\\\\ u|\_\{x=0\}=g(t),&t\geq 0 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解, 其中 \begin\{aligned\} \varphi(x),\quad \psi(x)=\left\\{\begin\{array\}\{ll\} 0,&0\leq x\leq 1,\\\\ \mbox\{正值\},&1 < x < \infty; \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} g(t)=\psi(x)=\left\\{\begin\{array\}\{ll\} 0,&0\leq x\leq 1,\\\\ \mbox\{正值\},&1 < x < \infty.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 试指出 $\displaystyle t > 0$ 时 $\displaystyle u(x,t)\equiv 0$ 的区域. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由初值条件的决定区域知当 \begin\{aligned\} at\leq x\leq 1-at, 0\leq t\leq \frac\{1\}\{2a\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 时, $\displaystyle u(x,t)=0$. 再由 d'Alembert 公式的几何意义知当 \begin\{aligned\} at\leq x\leq 1+at, 0\leq t\leq \frac\{1\}\{2a\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 时, $\displaystyle u(x,t)=0$. 其它的点由初边值即知 $\displaystyle u(x,t) > 0$. 于是 \begin\{aligned\} &\left\\{(x,t); u(x,t)=0\right\\}\\\\ =&\left\\{(x,t); at\leq x\leq 1-at, 0\leq t\leq \frac\{1\}\{2a\}\right\\}\\\\ &\cup \left\\{(x,t); at\leq x\leq 1+at, 0\leq t\leq \frac\{1\}\{2a\}\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 9、 求解达布 (Darboux) 问题 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} u\_\{tt\}-u\_\{xx\}=0, 0 < x < t, t > 0,\\\\ u|\_\{x=0\}=\varphi(t), u|\_\{x=t\}=\psi(t), t\geq 0, \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle \varphi(0)=\psi(0)$. 如果 $\displaystyle \varphi(t)$, $\displaystyle \psi(t)$ 都给定在 $\displaystyle [0,a]$ 上, 指出定解条件的影响区域. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 对在区域 $\displaystyle 0 < x < t$ 内的 $\displaystyle (x,t)$, 向下作特征线, 交 $\displaystyle x-t=0$ 于 \begin\{aligned\} \left(\frac\{x+t\}\{2\},\frac\{x+t\}\{2\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 交 $\displaystyle x=0$ 于 $\displaystyle (0,t-x)$. 找出平行四边形的另外一个顶点 \begin\{aligned\} \left(\frac\{t-x\}\{2\},\frac\{t-x\}\{2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而 \begin\{aligned\} u(x,t)=\varphi(t-x)+\psi\left(\frac\{x+t\}\{2\}\right)-\psi\left(\frac\{t-x\}\{2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 如果 $\displaystyle \varphi(t)$, $\displaystyle \psi(t)$ 都给定在 $\displaystyle [0,a]$ 上, 则由 d'Alembert 公式的几何意义知定解条件的影响区域为 \begin\{aligned\} \left\\{(x,t);\ x < t < x+\sqrt\{2\}a\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 事实上, 这个区域中的点向下作特征线, 会与 $\displaystyle [0,a]$ 上的初边值相交. 而这个区域外面的点向下作特征线, 不会与 $\displaystyle [0,a]$ 上的初边值相交. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 10、 设 $\displaystyle u(x,t)$ 是 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2u\_\{xx\}=0, x\in\mathbb\{R\}, t > 0,\\\\ u(x,0)=\varphi(x), u\_t(x,0)=\psi(x), x\in\mathbb\{R\} \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解, 并且对于 $\displaystyle |x|\geq 1, \varphi(x)=\psi(x)=0$. 证明: 对任意的 $\displaystyle x\_0$, 存在这样的数 $\displaystyle t\_0$ 与 $\displaystyle c$, 使得对所有的 $\displaystyle t\geq t\_0$, 有 $\displaystyle u(x\_0,t)=c$. 求出这些数. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 对 $\displaystyle \forall\ x\_0$, 取 $\displaystyle t\_0=\max\left\\{0,\frac\{1-x\_0\}\{a\},\frac\{1+x\_0\}\{a\}\right\\}$, 则当 $\displaystyle t\geq t\_0$ 时, $\displaystyle (x\_0,t)$ 的依赖区间 $\displaystyle [x\_0-at,x\_0+at]$ 与 $\displaystyle [-1,1]$ 没有交, 而 $\displaystyle u(x\_0,t)=0$. 故 $\displaystyle c=0$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 11、 设函数 $\displaystyle \varphi(x)\in C^2(\mathbb\{R\}), \psi(x)\in C^1(\mathbb\{R\})$, 且存在非零常数 $\displaystyle A,B$ 和 $\displaystyle b > 0$, 使得 \begin\{aligned\} \lim\_\{|x|\to+\infty\}|x|^\{-b\}\varphi(x)=A, \lim\_\{x\to+\infty\}|x|^\{1-b\}\psi(x)=B, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 证明一维波动方程的初值问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2u\_\{xx\}=0, x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=\varphi(x), u\_t|\_\{t=0\}=\psi(x), x\in\mathbb\{R\}\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解 $\displaystyle u(x,t)$ 满足: 存在常数 $\displaystyle C$, 使得 \begin\{aligned\} \lim\_\{t\to+\infty\}|t|^\{-b\}u(x,t)=C, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 并求出该常数 $\displaystyle C$. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} &|t|^\{-b\} u(x,t)=|t|^\{-b\}\frac\{\varphi(x-at)+\varphi(x+at)\}\{2\} +|t|^\{-b\} \frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau\\\\ =&\frac\{1\}\{2\}\left|\frac\{x\}\{t\}-a\right|^b \cdot |x-at|^\{-b\}\varphi(x-at) +\frac\{1\}\{2\}\left|\frac\{x\}\{t\}+a\right|^b \cdot|x+at|^\{-b\}\varphi(x+at)\\\\ &+|t|^\{1-b\} \frac\{1\}\{2at\}\int\_\{x-at\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau\\\\ \equiv&I\_1+I\_2+I\_3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 我们有 \begin\{aligned\} \lim\_\{t\to\infty\}I\_1=\frac\{1\}\{2\}a^b A, \lim\_\{t\to\infty\}I\_2=\frac\{1\}\{2\}a^b A. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 对 $\displaystyle I\_3$, 由 $\displaystyle \lim\_\{x\to+\infty\}|x|^\{-b\}\psi(x)=B$ 知 \begin\{aligned\} \forall\ \varepsilon > 0, \exists\ X > 0,\mathrm\{ s.t.\} \forall\ |x|\geq X, \left||x|^\{1-b\}\psi(x)-B\right| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是当 $\displaystyle t\geq \max\left\\{\frac\{x+X\}\{a\},\frac\{X-x\}\{a\}\right\\}$ 时, \begin\{aligned\} I\_3=&|t|^\{-b\} \frac\{1\}\{2a\}\left\[\int\_\{|\tau|\leq X\}\psi(\tau)\mathrm\{ d\} \tau +\int\_\{x-at\leq \tau\leq -X\atop X\leq \tau\leq x+at\}\psi(\tau)\mathrm\{ d\} \tau\right\]\\\\ \geq&\frac\{1\}\{2at^b\}\int\_\{|\tau|\leq X\}\psi(\tau)\mathrm\{ d\} \tau +\frac\{1\}\{2at^b\}\int\_\{x-at\leq \tau\leq -X\atop X\leq \tau\leq x+at\}\frac\{B-\beta\}\{|\tau|^\{1-b\}\}\mathrm\{ d\} \tau\\\\ =&\frac\{1\}\{2at^b\}\int\_\{|\tau|\leq X\}\psi(\tau)\mathrm\{ d\} \tau +\frac\{1\}\{2at^b\}\int\_X^\{at-x\}(B-\varepsilon)\tau^\{b-1\}\mathrm\{ d\} \tau\\\\ &+\frac\{1\}\{2at^b\}\int\_X^\{at+x\}(B-\varepsilon)\tau^\{b-1\}\mathrm\{ d\} \tau\\\\ =&\frac\{1\}\{2at^b\}\int\_\{|\tau|\leq X\}\psi(\tau)\mathrm\{ d\} \tau +\frac\{B-\varepsilon\}\{2ab\}\left\[\left(a-\frac\{x\}\{t\}\right)^b-\left(\frac\{X\}\{t\}\right)^b\right\]\\\\ &+\frac\{B-\varepsilon\}\{2ab\}\left\[\left(a+\frac\{x\}\{t\}\right)^b-\left(\frac\{X\}\{t\}\right)^b\right\] \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 同理, \begin\{aligned\} I\_3\leq&\frac\{1\}\{2at^b\}\int\_\{|\tau|\leq X\}\psi(\tau)\mathrm\{ d\} \tau +\frac\{B+\varepsilon\}\{2ab\}\left\[\left(a-\frac\{x\}\{t\}\right)^b-\left(\frac\{X\}\{t\}\right)^b\right\]\\\\ &+\frac\{B+\varepsilon\}\{2ab\}\left\[\left(a+\frac\{x\}\{t\}\right)^b-\left(\frac\{X\}\{t\}\right)^b\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 令 $\displaystyle t\to\infty$ 得 \begin\{aligned\} \frac\{B-\varepsilon\}\{ab\}a^b\leq \varliminf\_\{t\to\infty\}I\_3\leq \varlimsup\_\{t\to\infty\}I\_3\leq \frac\{B+\varepsilon\}\{ab\}a^b. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 再令 $\displaystyle \varepsilon\to 0^+$ 得 $\displaystyle \lim\_\{t\to\infty\}I\_3=\frac\{Ba^\{b-1\}\}\{b\}$. 于是 \begin\{aligned\} \lim\_\{t\to\infty\}t^\{-b\}u(x,t)=Aa^b+\frac\{Ba^\{b-1\}\}\{b\}\equiv C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 12、 求解下列非齐次方程的定解问题: (1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-u\_\{xx\}=t\sin x,&x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=0, u\_t|\_\{t=0\}=\sin x,&x\in\mathbb\{R\};\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle a=1, f(x,t)=t\sin x, \phi(x)=0, \psi(x)=\sin x$, 则 \begin\{aligned\} u(x,t)=&\frac\{\phi(x-at)+\phi(x+at)\}\{2\}+\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau\\\\ &+\frac\{1\}\{2a\}\int\_0^t \mathrm\{ d\} \tau \int\_\{x-a(t-\tau)\}^\{x+a(t-\tau)\}f(\xi,\tau)\mathrm\{ d\} \xi\\\\ =&0+\sin t\sin x+(t-\sin t)\sin x=t\sin x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-4u\_\{xx\}=2,&x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=x^2, u\_t|\_\{t=0\}=0,&x\in\mathbb\{R\};\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle a=2, f(x,t)=2, \phi(x)=x^2, \psi(x)=0$, 则 \begin\{aligned\} u(x,t)=&\frac\{\phi(x-at)+\phi(x+at)\}\{2\}+\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau\\\\ &+\frac\{1\}\{2a\}\int\_0^t \mathrm\{ d\} \tau \int\_\{x-a(t-\tau)\}^\{x+a(t-\tau)\}f(\xi,\tau)\mathrm\{ d\} \xi\\\\ =&(4t^2+x^2)+0+t^2=5t^2+x^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-9u\_\{xx\}=120t,&x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=x^2, u\_t|\_\{t=0\}=0,&x\in\mathbb\{R\}.\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle a=3, f(x,t)=120t, \phi(x)=x^2, \psi(x)=0$, 则 \begin\{aligned\} u(x,t)=&\frac\{\phi(x-at)+\phi(x+at)\}\{2\}+\frac\{1\}\{2a\}\int\_\{x-at\}^\{x+at\}\psi(\tau)\mathrm\{ d\} \tau\\\\ &+\frac\{1\}\{2a\}\int\_0^t \mathrm\{ d\} \tau \int\_\{x-a(t-\tau)\}^\{x+a(t-\tau)\}f(\xi,\tau)\mathrm\{ d\} \xi\\\\ =&(9t^2+x^2)+0+20t^3=t^2(20t+9)+x^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/